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3 years ago

Electromagnetic waves differ only in their _____, their energy, and their frequency.

Physics

1 answer:

Tanya [424]3 years ago

6 0

Answer:

Wavelength

Explanation:

Electromagnetic Waves have different wavelengths.

They only differ from each other in wavelength. Wavelength is the distance between one wave crest to the next.

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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

6 0

3 years ago

What tripositive ion has the electron configuration [kr] 4d3 ? what neutral atom has the electron configuration [kr] 5s 2 4d2 ?

tigry1 [53]

(a)

Electronic configuration is given as follows:

$[Kr]4d^{3}$

Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.

Thus, electronic configuration of neutral atom is $[Kr]4d^{5}5s^{1}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.

This corresponds to element: molybdenum. Thus, the tripositive atom will be $Mo^{3+}$ .

(b) The given electronic configuration is $[Kr]5s^{2}4d^{2}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.

This corresponds to element zirconium, represented by symbol Zr.

6 0

3 years ago

A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a

aleksklad [387]

Answer:

a) y = 0.98 t², t=1s y= 0.98 m,

b) he two blocks must move the same distance

c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

N-W = 0

N = M g

X axis

T- fr = Ma

the friction force has the expression

fr = μ N

fr = μ Mg

small block

w- T = m a

we write the system of equations

T - fr = M a

mg - T = m a

we add and resolved

mg- μ Mg = (M + m) a

a = $g \ \frac{m - \mu M}{m+M}$

a = $9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}$

a = 9.8 (6/30)

a = 1.96 m / s²

a) now we can use the kinematic relations

y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

y = ½ a t²

y = ½ 1.96 t²

y = 0.98 t²

for t = 1s y = 0.98 m

t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

v = vo + a t

v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

-fr = M a

- μ M g = M a

a = - μ g

a = - 0.2 9.8

a = -1.96 m / s²

e) the distance to stop the block is

v² = vo² - 2 a x

0 = vo² - 2a x

x = vo² / 2a

x = 1.96² / 2 1.96

x = 0.98 m

the time to travel this distance is

v = vo - a t

t = vo / a

t = 1.96 /1.96

t = 1 s

3 0

3 years ago

What is the kinetic energy of a 1.40 kg discus with a speed of 22.5 m/s?

Oksana_A [137]

Kinetic energy = (1/2) (mass) (speed)²

   = (1/2) (1.4 kg) (22.5 m/s)²

   = (0.7 kg) (506.25 m²/s² )

   = 354.375 kg-m²/s² = 354.375 joules .

This is just the kinetic energy associated with a 1.4-kg glob of
mass sailing through space at 22.5 m/s. In the case of a frisbee,
it's also spinning, and there's some additional kinetic energy stored
in the spin.

8 0

3 years ago

This is a voice recording of the text of a book that you listen rather than read.

OleMash [197]

Romeo and Juliet is one

8 0

3 years ago